User mikhail bondarko - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:14:22Z http://mathoverflow.net/feeds/user/2191 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46751/does-there-exist-a-functorial-splitting-for-the-weight-filtration-of-singular-co Does there exist a functorial splitting for the weight filtration (of singular cohomology)? Mikhail Bondarko 2010-11-20T17:13:12Z 2013-04-23T01:42:52Z <p>There are plenty of examples of varieties whose singular cohomology with rational coefficients considered as a mixed Hodge structure does not decompose as the direct sum of its pure (weight) factors. Yet if we consider the cohomology just as filtered vector spaces over rationals, such a decomposition certainly exist (for any variety). </p> <p>My question is: could there exist a functorial decomposition like this (say, for the singular cohomology as a functor from the category of all smooth complex varieties), or does there exist some obstruction for such a functorial splitting?</p> http://mathoverflow.net/questions/127275/what-is-the-purpose-of-section-3-of-bbd What is the purpose of section 3 of BBD? Mikhail Bondarko 2013-04-11T19:05:48Z 2013-04-12T02:12:03Z <p>I am not quite sure that this question is appropriate for Mathoverflow, yet I would be deeply grateful for any hint: what happens in section 3 of Beilinson A., Bernstein J., Deligne P., Faisceaux pervers//Asterisque 100, 1982, 5-171? Is there any statement that is important for the following sections of this treatise? I do not know French; yet this does not prevent me from understanding all the other sections.</p> <p>Upd. I wonder: does there exist a 'plan' of BBD?</p> http://mathoverflow.net/questions/126817/how-would-you-say-that-a-small-category-is-embedded-into-functors-from-a-large-c How would you say that a small category is embedded into functors from a large $C'$ to abelian groups? Mikhail Bondarko 2013-04-08T02:52:21Z 2013-04-08T05:00:24Z <p>How would you say that a small additive category $C$ embedds (contravariantly) into the category of exact functors from a 'large' abelian $C'$ into abelian groups (this is something like Yoneda's embedding, but $C$ does not map canonically into $C'$)? My problem is that I do not want to consider all functors from $C'$ into abelian groups since this functor category it 'very large' (and I do not want to consider a 'larger universe'). Certainly, I can try to consider a limit of the corresponding functors from small subcategories of $C'$; yet is there a better way to deal with this matters?</p> <p>Upd. Actually, my $C'$ is just isomorphic to the category of additive functors from $C^{op}$ to abelian groups (though this is not the way how it is defined).</p> http://mathoverflow.net/questions/126693/singularity-locus-in-terms-of-ideals Singularity locus in terms of ideals. Mikhail Bondarko 2013-04-06T06:41:37Z 2013-04-06T06:41:37Z <p>Let $X$ be a smooth affine variety over a field, $Z\subset Y\subset X$ are closed (reduced) subvarieties. What are the possible ways to verify whether $Y$ is singular at $Z$ i.e. whether $Z$ is contained in the singularity locus of $Y$? I would prefer some conditions in terms of ideals that determine $Y$ and $Z$ in $X$; yet other criteria could also be useful.</p> http://mathoverflow.net/questions/125469/when-r-f-is-regular When $R/(f)$ is regular? Mikhail Bondarko 2013-03-24T16:57:58Z 2013-03-25T11:43:12Z <p>For R being a commutative regular excellent Noetherian ring of finite Krull dimension which conditions on $f\in R$ can ensure that the ring $R/(f)$ is regular (so, I want a sufficient condition)? I do not want to look at all points of $R$.</p> <p>Upd1. My $R$ is the inductive limit of a system of inclusions of regular Noetherian rings $R_i$ of finite Krull dimension. So, I want a 'finite number of conditions' on $f$ since I would like to check them for $R_i$. </p> <p>Upd2. Since I am interested in algebras over fields, it seems that regularity can be characterized in terms of Andre-Quillen homology. Yet my algebras are not of finite type; are there any 'finite' substitute for these homology groups that ensure regularity (I do not need a necessary and sufficient criterion). </p> http://mathoverflow.net/questions/124209/regular-subscheme-of-a-projective-limit-of-schemes Regular subscheme of a projective limit of schemes Mikhail Bondarko 2013-03-11T08:42:22Z 2013-03-11T08:42:22Z <p>Let $S\cong \varprojlim S_i$, where $S$ and all $S_i$ are separated regular excellent of finite Krull dimension. Let $Z$ be a closed regular subscheme of $S$. As Theorem 8.8.2 of EGA4 shows, $Z$ comes from a closed subscheme $Z_i$ of one of $S_i$. My question is: can we also assume that $Z_i$ is regular? Are there any extra restrictions needed so that the fact will be true? I am actually only interested in equicharacteristic schemes, and the connecting morphisms are affine and dominant. </p> http://mathoverflow.net/questions/123716/which-schemes-can-be-presented-as-limits-of-smooth-varieties Which schemes can be presented as limits of smooth varieties? Mikhail Bondarko 2013-03-06T07:17:36Z 2013-03-06T10:06:52Z <p>I can prove a certain statement for any scheme that can be presented as the limit of an essentially affine (filtering) projective system of smooth varieties over a perfect field such the connecting morphisms are dominant. In this text I only treat schemes that are excellent separated of finite Krull dimension. So, I have the following questions.</p> <ol> <li><p>Is there a shorter description of schemes that can be presented as limits of this sort (either of all ones, or of excellent separated of finite Krull dimension)?</p></li> <li><p>Is there an interesting subclass in the class of all limit schemes of this sort? I don't want to restrict myself to affine schemes.</p></li> <li><p>If no nice answers to questions 1 and 2 will be given, I will need a term for limits of this sort. Any suggestions?:)</p></li> </ol> http://mathoverflow.net/questions/120553/should-the-etale-cohomology-of-a-smooth-projective-variety-over-rationals-be-se Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why? Mikhail Bondarko 2013-02-01T21:53:27Z 2013-02-27T15:16:48Z <p>$\DeclareMathOperator{\char}{char}\DeclareMathOperator{\gal}{Gal}$ Let $P$ be a smooth projective variety over a field $K$ (one may certainly assume that $K$ is perfect; the case $K=\mathbb{Q}$ already seems to be interesting enough). For some $\ell\neq \char K$, $n>0$, should the $n$-th $\mathbb Q_\ell$-adic Galois cohomology of $X_{K^{sep}}$ be semi-simple as a $\gal(K)$-representation? Certainly; no proof of this fact is known, so I would rather like to know whether it is related with some 'motivic' conjectures.</p> <p>Some remarks:</p> <ol> <li><p>For a finite $K$ one can consider the 'motivic' Frobenius; thus the conjecture follows from standard (motivic) ones. Yet this argument does not seem to work already for $K=\mathbb Q$.</p></li> <li><p>It is certainly tempting to apply some polarizability argument. Yet my impression is that polarizability can only be applied to Hodge structures (in general); is this true?</p></li> </ol> <p>Upd. It seems (see the comment of Ulrich) that 'my conjecture' is wrong for $K= \mathbb Q_\ell$; this settles my question. Yet I wonder where I can find the details for this example (when is the representation corresponding to an elliptic curve with multiplicative reduction indecomposable). </p> http://mathoverflow.net/questions/122697/how-would-you-call-a-subscheme-of-a-smooth-s-scheme How would you call a subscheme of a smooth $S$-scheme? Mikhail Bondarko 2013-02-23T06:32:10Z 2013-02-23T08:35:58Z <p>In my preprint I propose to call $X/S$ quasi-smooth if $X$ can be embedded into a smooth $X'/S$. Does this sound fine?</p> <p>Upd. So, smoothly embeddable is better? Is it ok to call a morphism smoothly embeddable?</p> http://mathoverflow.net/questions/122556/if-x-y-are-regular-and-of-finite-type-over-s-can-x-times-s-y-be-embedded If $X,Y$ are regular and of finite type over $S$, can $X\times _S Y$ be embedded into a regular $S$-scheme? Mikhail Bondarko 2013-02-21T17:37:15Z 2013-02-21T21:29:59Z <p>It seems to be well-known (see <a href="http://mathoverflow.net/questions/201/is-there-an-example-of-a-variety-over-the-complex-numbers-with-no-embedding-into" rel="nofollow">http://mathoverflow.net/questions/201/is-there-an-example-of-a-variety-over-the-complex-numbers-with-no-embedding-into</a>) that a general finite type $S$-scheme does not embedd into a regular $S$-one even when $S$ is (the spectrum of) a field. Now, I am interested in the following setting: $X,Y$ are regular schemes of finite type over $S$; $S$ is separated excellent noetherian of finite Krull dimension (and one may assume that $X$ and $Y$ are proper over it). Then is it true that $X\times _S Y$ (or the corresponding reduced scheme) can be embedded into a regular scheme of finite type over $S$? </p> <p>This is certainly true if $X,Y$ are quasi-projective over $S$ (so the question is not 'local'). On the other hand, if $S$ is smooth over a characteristic zero field $K$, then $X,Y$ are smooth over $K$, and $X\times_S Y\subset X\times_K Y$ (we can map the product to $S$ via the first component). </p> <p>So, I wonder for what $S$ such an embedding always exists; I would be grateful for any hints or evidence! </p> http://mathoverflow.net/questions/121996/applications-for-intersection-cohomology-and-for-the-decomposition-theorem-for Applications for intersection (co)homology and for the Decomposition Theorem for students? Mikhail Bondarko 2013-02-16T16:24:07Z 2013-02-17T17:07:32Z <p>Which applications of intersection (co)homology and of the (Topological) Decomposition Theorem have most chances to be understood by students?</p> http://mathoverflow.net/questions/119914/a-canonical-way-to-kill-a-subset-of-cohomology-in-a-dg-algebra-via-a-infty-al A canonical way to kill a subset of cohomology in a dg-algebra: via $A_\infty$-algebras? References? Mikhail Bondarko 2013-01-26T04:41:28Z 2013-01-28T06:57:28Z <p>Let $A$ be a differential graded algebra, $S\subset H^*(A)$. I would like to 'kill $S$ in a canonical way'. Is it possible to do it as follows: consider the $A_\infty$-algebra structure on $H^\ast(A)$, and factorize $H^*(A)$ by the '$A_\infty$-ideal' generated by $S$ (i.e. kill in $H^\ast(A)$ all sums of elements that can be obtained via $m_i$ taking elements of $S$ as one of the arguments)? Does this quotient possess any nice 'universal' properties? What other reasonable definitions of $H^\ast(A)/\langle S\rangle$ can one consider here? </p> <p>I would like to concider the (triangulated?) category of $A_\infty$-modules over $H^\ast(A)/\langle S\rangle$; is there a way to describe it 'explicitly'? What is the relation between $A$-modules and $H^\ast(A)/\langle S\rangle$-ones? Can one define a functor by $A-\mod\cong H^\ast(A)-\mod \stackrel{\otimes_ {H^\ast(A)}H^\ast(A)/\langle S\rangle}{\to} H^\ast(A)/\langle S\rangle-mod$; does it possess any nice universal properties?</p> <p>What are the 'canonical' references for these matters (in particular, for factorizing $A_\infty$-algebras modulo ideals)? </p> <p>Upd. Unfortunately, I would like to consider an algebra that is not skew-commutative; it is closely related with a 'complicated' DG-category.</p> http://mathoverflow.net/questions/119426/ring-structure-for-the-motivic-spectrum-complex-that-represents-singular-cohomolo Ring structure for the motivic spectrum/complex that represents singular cohomology? Mikhail Bondarko 2013-01-20T21:24:36Z 2013-01-20T23:38:12Z <p>As the discussion here <a href="http://mathoverflow.net/questions/42693/is-singular-cohomology-representable-by-a-voevodskys-motivic-complex" rel="nofollow">http://mathoverflow.net/questions/42693/is-singular-cohomology-representable-by-a-voevodskys-motivic-complex</a> shows, the singular cohomology of (smooth) complex varieties is represented by a motivic complex (and also by a motivic spectrum). My question is: what can be proved about the ring structure for this complex/spectrum? Is it known that this a 'weak' ring spectrum? an $A_{\infty}$-spectrum? a highly structured ring spectrum? Any hints would be very welcome!</p> http://mathoverflow.net/questions/118867/a-certain-coniveau-like-filtration-for-cohomology-what-can-one-say-about-the-i A certain 'coniveau-like' filtration for cohomology: what can one say about the intersection of $Ker H^i(X)\to H^i(Z)$ for $Z$ running through subvarieties of $X$ of dimension $m$? Mikhail Bondarko 2013-01-14T10:55:22Z 2013-01-14T10:55:22Z <p>Let $X$ be a smooth variety (say, a complex one; denote its dimension by $n$). What can one say about the intersection of $Ker (H^i(X)\to H^i(Z))$ for $Z$ running through (closed, not necessarily smooth!!) subvarieties of $X$ of dimension $m$ (here $H^\ast$ is singular or etale cohomology)? Did anybody study this filtration (for fixed $X,i$, when $m$ varies) before? What properties can one prove for it?</p> <p>Some remarks.</p> <ol> <li><p>If $Z$ is a generic multiple hyperplane section of $X$, $i\le m$, then a theorem of Beilinson yields that the map $H^i(X)\to H^i(Z)$ is injective.</p></li> <li><p>On the other hand, I believe that $H^{m+1}(X)\to H^{m+1}(Z)$ cannot be injective if $X$ is 'too complicated'; yet I have no idea how to construct any more or less general examples here.</p></li> <li><p>Certainly, it suffices to consider only 'large enough' $Z$ here.</p></li> <li><p>The filtration in question is closely related with the one given by $\cap Ker (H^i(f):H^i(X)\to H^i(Z))$ for $f$ running through proper morphisms with $Z$ of dimension $m$ (and smooth).</p></li> <li><p>If $X$ is proper, then one can replace $H^\ast$ here with $H_c^\ast$. </p></li> </ol> http://mathoverflow.net/questions/118663/for-which-local-r-its-k-theory-mod-l-is-isomorphic-to-the-one-of-its-residue-fi For which local $R$ its K-theory mod l is isomorphic to the one of its residue field? Mikhail Bondarko 2013-01-11T18:54:21Z 2013-01-12T21:00:55Z <p>It is well-known (and was proved by Gabber?): if $R$ is a regular henselian local ring containing a field of characteristic prime to $l$, $k$ is its residue field, then $K_\ast(R,\mathbb{Z}/l)\cong K_\ast(k,\mathbb{Z}/l)$. My question is: are there any more classes of (regular) local rings such that this is true for them? Conversely, for which types of local rings this statement is 'usually' wrong?</p> http://mathoverflow.net/questions/118708/primitive-cohomology-useful/118739#118739 Answer by Mikhail Bondarko for Primitive Cohomology Useful? Mikhail Bondarko 2013-01-12T16:19:35Z 2013-01-12T16:19:35Z <p>One more application: the singular cohomology functor (with coefficients in a field) restricted to smooth projective complex varieties factorizes through the semi-simple category of polarizable pure Hodge structures. There is a certain (somewhat complicated) extension of this result to cohomology of all complex varieties.</p> http://mathoverflow.net/questions/118437/some-weak-proper-and-smooth-base-change-theorems-for-nisnevich-sheaves Some 'weak proper and smooth base change' theorems for Nisnevich sheaves? Mikhail Bondarko 2013-01-09T10:27:20Z 2013-01-09T10:27:20Z <p>Among the most important tools for studying etale cohomology are the proper and smooth base change theorems. I suspect that these theorems are no longer true for Nisnevich cohomology (probably finite morphisms of fields may already give counterexamples). Yet are there some classes of morphisms of varieties for which certain Nisnevich analogues of the base change theorems are known?</p> http://mathoverflow.net/questions/117942/the-fibres-of-smooth-projective-families-over-all-geometric-points-have-isomorphi The fibres of smooth projective families over all geometric points have isomorphic cohomology; are these isomorphisms 'functorial'? Mikhail Bondarko 2013-01-03T11:06:16Z 2013-01-06T03:32:32Z <p>Let $p:P\to S$ (and $p':P\to S$) be proper smooth morphisms of 'nice' schemes (one may assume that $S$ is a complex variety). It is well-known that the fibres of $p$ (and $p'$) over all geometric points of $S$ have isomorphic etale cohomology. Now, let $f:P\to P'$ be an $S$-morphism (that is certainly proper, but not necessarily smooth). My question is: is $f$ necessarily compatible with the isomorphisms mentioned?</p> <p>It suffices to consides two points $s_{0,1}$ of $S$, $s_1$ is a generic point. As far as I understand the proof, the isomorphisms mentioned are obtained by applying cohomology to the base change via $p$ and $p'$ of the natural morphisms $s_i\to Spec O^h_{s_0}$ ($i=0,1$, $s_i$ are the corresponding spectra of fields, $O^h_{s_0}$ is the henselization of $S$ at $s_0$). If this is correct, then the answer to my question should be positive for obvious reasons. I have serious doubts here since this seems to contradict some other well known facts.</p> <p>Any hints or references would be very welcome!</p> http://mathoverflow.net/questions/117099/a-nice-way-to-verify-whether-the-neron-severi-group-of-a-smooth-affine-variety-is A nice way to verify whether the Neron-Severi group of a smooth affine variety is zero Mikhail Bondarko 2012-12-23T18:19:55Z 2012-12-23T18:19:55Z <p>Let $S$ be a smooth affine variety over an algebraically closed field (this could be the field of complex numbers). Is there an 'easy' way to verify whether $NS(S)=0$? Unfortunately, I don't know how to compute the cohomology of my $S$; I only know that it has 'many holes' in it (i.e. it is 'very far from being projective'), and it is finite over a variety $X$ with $NS(X)=0$ and 'very many holes'.</p> <p>Any advice would be very welcome! In particular, are there any special methods that can simplify the calculation of $H^2(S)$ in this situation?</p> http://mathoverflow.net/questions/115953/picx-l0-in-terms-of-h-etx-mu-ln $Pic(X)/l=0$ in terms of $H^*_{et}(X,\mu_{l^n})$? Mikhail Bondarko 2012-12-10T06:02:05Z 2012-12-10T17:31:05Z <p>I would like to calculate Picard groups of certain schemes over fields; I'm mostly interested in the question whether $Pic(X)$ is infinitely $l$-divisible, i.e. whether $Pic(X)/l=0$, $l$ is a prime distinct from the base field characteristic (the latter could be $0$). I would like to have a characterization of this vanishing in terms of etale cohomology of $l$-torsion sheaves.</p> <p>Certainly, $Pic(X)\cong H^1_{et}(X,G_m)$; yet this only yields an injection of $Pic(X)/l$ into $H^1(X,\mu_l)$, and I don't know how to control the cokernel.</p> <p>Upd. So, is there a general method that expresses $Pic(X)/l$ in terms of all of $H^i(X,\mu_{l^n})$ (for $i,n>0$)? What can be said here if $X$ is a variety over an algebraically closed field?</p> http://mathoverflow.net/questions/114193/voevodskys-split-standard-triple-argument-an-explanation-does-it-work-with Voevodsky's 'split standard triple' argument: an explanation; does it work with $Z/nZ$-coefficients? Mikhail Bondarko 2012-11-22T22:42:24Z 2012-11-26T09:14:43Z <p>For varieties over a perfect field (of some characteristic $p$ that could be 0) Voevodsky defines the notion of a 'standard triple' (see <a href="http://books.google.ru/books?id=TzUmk87bN9cC&amp;pg=PA85&amp;lpg=PA85&amp;dq=Voevodsky+standard+triple&amp;source=bl&amp;ots=lqFZojfUU-&amp;sig=Jtt57xtmlQwX7XvShXHPyaKVP68&amp;hl=ru&amp;sa=X&amp;ei=SaeuUIzEGZH24QSEwIG4DA&amp;redir_esc=y#v=onepage&amp;q=Voevodsky%20standard%20triple&amp;f=false" rel="nofollow">http://books.google.ru/books?id=TzUmk87bN9cC&amp;pg=PA85&amp;lpg=PA85&amp;dq=Voevodsky+standard+triple&amp;source=bl&amp;ots=lqFZojfUU-&amp;sig=Jtt57xtmlQwX7XvShXHPyaKVP68&amp;hl=ru&amp;sa=X&amp;ei=SaeuUIzEGZH24QSEwIG4DA&amp;redir_esc=y#v=onepage&amp;q=Voevodsky%20standard%20triple&amp;f=false</a>). One says that a triple is split over $U$ if a certain line bundle is trivial (see Definition 11.11); this has certain consequences for cohomology of varieties with coefficients in a homotopy invariant presheaf with transfers $F$ (see Proposition 11.15).</p> <p>My question is: if $nF=0$, is it sufficient to consider triviality modulo $n$ instead, i.e. could one replace all the Picard groups considered in this section by their $\mathbb{Z}/n\mathbb{Z}$-analogues? I looked at the proofs, and it seems that the answer is positive; yet possibly I miss something.</p> <p>Alternatively, one can find Voevodsky's (Mazza's-Weibel's) book here <a href="http://www.claymath.org/library/monographs/cmim02c.pdf" rel="nofollow">http://www.claymath.org/library/monographs/cmim02c.pdf</a> an earlier exposition of this argument can be found in section 4 of <a href="http://www.math.illinois.edu/K-theory/0368/s3.pdf" rel="nofollow">http://www.math.illinois.edu/K-theory/0368/s3.pdf</a></p> <p>Upd. Possibly, a more clear reference to Voevodsky's argument is <a href="http://www.math.uiuc.edu/K-theory/0832/motvo.pdf" rel="nofollow">http://www.math.uiuc.edu/K-theory/0832/motvo.pdf</a>, section 5.1.1; yet I would be deeply grateful for any 'explanation' of this reasoning. </p> http://mathoverflow.net/questions/114177/for-a-finite-flat-etale-morphism-fy-to-x-is-f-1-y-deg-f-1-x-nilpoten For a finite flat (etale?) morphism $f:Y\to X$, is $f_*1_Y-\deg f . 1_X$ nilpotent in $A^0(X)$, where $A^*$ is the algebraic cobordism? Mikhail Bondarko 2012-11-22T16:28:35Z 2012-11-23T19:38:33Z <p>Let $A^\ast$ be an algebraic oriented cohomology theory (i.e. it is equipped with certain push-forwards for projective morphisms of smooth varieties over the base field $k$; see section 2 of <a href="http://www.math.uiuc.edu/K-theory/0535/orient.pdf" rel="nofollow">http://www.math.uiuc.edu/K-theory/0535/orient.pdf</a> for more detail); let $f:Y\to X$ be a finite morphism of smooth varieties whose degree is $d$. I would like to prove the following conjecture: if $f^\ast h=0$ for $h\in A^\ast (X)$, then $d^lh=0$ for some $l>0$ (one cannot take $l=1$ here when $A^*$ is the K-theory). To this end it suffices to verify that $f_\ast 1_Y-d$ is nilpotent in $A^0(X)$ (since $f_\ast f^\ast h=f_\ast 1_Y\cdot h$ by the property (v) in the reference cited). It seems sufficient to prove the latter for $A^\ast$ being the algebraic cobordism (as defined by Levine and Morel), since this is the universal algebraic oriented cohomology theory.</p> <p>I would like to say that $f_\ast 1_Y-d$ is supported in codimension 1. Does $A^0(X)$ possess a multiplicative coniveau filtration? If $f$ is generically etale, then I can use the fact that $f'_{*}1_{Y'}=d$ for $f':Y'\to X'$ being the (etale) morphism of generic points; Levine proves this in his cobordism book. </p> <p>Is there a better way to prove my conjecture (that would work even if $f$ is not generically etale)?</p> http://mathoverflow.net/questions/114016/exceptional-collections-of-objects-in-topological-triangulated-categories Exceptional collections of objects in topological triangulated categories? Mikhail Bondarko 2012-11-21T03:39:20Z 2012-11-21T20:50:34Z <p>People often consider exceptional sets of objects (i.e. collections of objects satisfying certain strong orthogonality conditions: $Ext^{l}(P_i,P_j)$ should be zero for $l\neq 0$ + something else) in derived categories of coherent sheaves (over algebraic varieties; possibly the first example corresponds to the Beilnson's description of the derived category of coherent sheaves on the projective space of dimension $n$). Are there any examples of this notion in some stable homotopy categories (in the sense of abstract model categories; one can consider the category of modules over a ring spectrum here)?</p> http://mathoverflow.net/questions/112551/if-a-t-truncation-of-the-unit-object-in-a-stable-homotopy-category-is-a-ring-obje If a t-truncation of the unit object in a stable homotopy category is a ring object up to homotopy, can it be lifted to a ring spectrum? What about the Postnikov t-truncations of the sphere spectrum? Mikhail Bondarko 2012-11-16T04:47:34Z 2012-11-16T14:48:42Z <p>Let $S$ be the unit object in a monoidal stable homotopy category $SH$ (we demand that the multiplication $S\times S\to S$ is commutative and associative on the level of spectra, and not just up to weak equivalence). Let $\tau$ be a $t$-structure for $SH$ such that $S$ is $\tau$-negative and that $SH^{\tau\le 0}\times SH^{\tau\le 0}\subset SH^{\tau\le 0}$. Then for any $m\le 0$ the object $S'=S^{\tau \ge m}$ (a 'factor' of $S$) can be easily seen to be a monoidal object in $SH$. My question is: can $S'$ be lifted to a commutative ring spectrum? Which restrictions on $SH,S,t$ are needed to do this? Are there any other ways of 'rigidifying' $S'$ such that one can still consider a certain triangulated category of $S'$-modules?</p> <p>Basically I am interested in 'motivic' stable homotopy categories ($SH$, $MGL$-modules, and 'big Voevodsky's motives'), yet the easiest examples of my setting are the Postnikov $t$-truncations of the sphere spectrum in the 'usual' ('topological') $SH$ (for $m=0$ this is the Eilenberg-Maclane spectrum for $\mathbb{Z}$). Also, what can be said about the complex cobordism spectrum and modules over it?</p> http://mathoverflow.net/questions/112352/interesting-examples-of-a-4-torsion-x-in-a-triangulated-category-such-that-2-end Interesting examples of a 4-torsion X in a triangulated category such that $2 End(X/2X)\neq 0$? Mikhail Bondarko 2012-11-14T06:14:00Z 2012-11-14T14:11:42Z <p>It is well-known that for the sphere spectrum $S$ in the ('topological') stable homotopy category the object $S/2S$ i.e. the cone of $S\stackrel{\times 2}{\to}S$, is not $2$-torsion. </p> <p>So I wonder where there exists an object $X$ in a (topological?) triangulated category such that</p> <ol> <li><p>$2End(X/2X)\neq 0$.</p></li> <li><p>$End(X,X)$ is torsion ($\cong \mathbb{Z}/4 \mathbb{Z}$?).</p></li> <li><p>$Hom(X,X[i])=0$ for any $i\neq 0$ (or at least for 'small' $i$). </p></li> </ol> <p>I would be grateful for any hints or references concerning this question! I believe that I have proved that condition 3 contradicts 2 if $End(X,X)\cong \mathbb{Z}$ (since in this case the triangulated subcategory 'strongly' generated by $X$ is isomorphic to $K^b(B)$, where $B$ is the category of finitely generated free $\mathbb{Z}$-modules); yet I cannot prove anything like that if condition 2 is fulfilled.</p> http://mathoverflow.net/questions/111941/good-morphisms-of-distinguished-triangles-can-neemans-method-be-applied-to-the Good morphisms of distinguished triangles: can Neeman's method be applied to the motivic stable homotopy category? Mikhail Bondarko 2012-11-09T21:27:25Z 2012-11-11T21:05:14Z <p>It is well known that non-uniqueness of a cone for a morphism in a triangulated category $C$ makes constructing exact functors (of triangulated categories) a difficult task. In section 3 of his "Some new axioms for triangulated categories" Neeman proposes an alternative axiomatics of triangulated categories (that includes a certain notion of 'good morphism' of distinguished triangles). I wonder whether Neeman's methods can be applied to the motivic stable homotopy category $SH$ of Morel and Voevodsky? Can any better methods for 'rigidifying' (for example, higher categories?) be applied to $SH$? </p> <p>Upd. It seems that I need an category $G$ whose objects and morphism are morally distinguished triangles in $C$ and their morphisms + some extra data. Any $C$-morphism can be extended to an object of $G$, any commutative square could be extended to a $G$-morphism, whereas for any $G$-morphism of distinguished triangles $X\to Y\stackrel{v}{\to} Z\stackrel{w}{\to} X[1]$ and $X'\to Y'\stackrel{v}{\to} Z'\stackrel{w'}{\to} X'[1]$ that is $0$ on $X$ and $Y$ the corresponinding morphism from $Z$ to $Z'$ equals $v'\circ \theta \circ w$ for some $\theta\in C(X[1],Y')$ (this is the correction of the Neeman's axiom GTR2).</p> <p>Denis-Charles Cisinski has written some very interesting comments on the relations between various types of 'nice' triangulated categories; yet I wonder which texts treat this topic. </p> http://mathoverflow.net/questions/111849/the-vanishing-of-mgl2ni-nx-do-spectra-of-smooth-projective-varieties-gen The vanishing of $MGL^{2n+i,n}(X)$; do spectra of smooth projective varieties generate $SH_{l}$? Mikhail Bondarko 2012-11-08T20:53:08Z 2012-11-11T07:05:54Z <p>I have two questions related to the stable motivic homotopy categories of Morel-Voevodsky. The first is probably simple; I wonder what is known on the second one.</p> <ol> <li><p>For the algebraic cobordism theory $MGL$ and a smooth variety $X$ over a (perfect?) field is it true that $MGL^{2n+i,n}(X)=0$ for any $n\in \mathbb{Z},i>0$? More generally, are there any reasonable restrictions on a (oriented?) ring spectrum $E$ in $SH$ that ensure the vanishing of<br> $MGL^{2n+i,n}(E)$. In particular, is this question related with some sort of effectivity for spectra? </p></li> <li><p>It is well known that 'shifts and twists' of the spectra $\Sigma(X_+)$ generate $SH$, where $X$ runs through all smooth $k$-varieties. If the characteristic of $k$ is $0$, resolution of singularities yields that it suffices to consider only smooth projective varieties here. Now, what statements of this sort are known for $k$ of characteristic $p>0$? I suspect that that one can deduce a similar result for $SH\otimes \mathbb{Z}_{(l)}$ for any prime $l\neq p$, ffrom the Gabber's l'-alterations theorem. Is this true? If this is too difficult, can one prove a similar statement for the triangulated category of $MGL$-modules? </p></li> </ol> <p>What are the best references for these questions?</p> http://mathoverflow.net/questions/111579/how-can-one-compute-the-cohomology-of-ic-for-i-mathbban-1-setminus How can one compute the cohomology of $i'^*C$, for $i':\mathbb{A}^{N-1}\setminus \{0\}\to \mathbb{A}^{N}\setminus \{0\}$? Mikhail Bondarko 2012-11-05T20:56:37Z 2012-11-05T20:56:37Z <p>For an (etale or 'topological', constructible bounded) complex of sheaves $C$ on $X'=\mathbb{A}^{N}\setminus {0}$, $i'$ being the embedding $\mathbb{A}^{N-1}\setminus {0}\to \mathbb{A}^{N}\setminus {0}$ I would like to compute the cohomology of $Ri'_*C$. Are there any 'nice' ways to do this? I would to consider the embedding $j: \mathbb{A}^{N}\setminus {0} \to \mathbb{A}^{N}$ and restrict $Rj_*C$ to $\mathbb{A}^{N-1}$ instead; yet we do not have a base change isomorphism here, and so the cohomology of $Ri^*Rj_*C$ is not very much related with the one of $Ri'_*C$. Is there any way to 'fix' this nuisance?</p> http://mathoverflow.net/questions/111353/can-there-exist-chow-motives-motivic-cohomology-for-compact-kahler-manifolds Can there exist Chow motives/motivic cohomology for compact Kähler manifolds? Mikhail Bondarko 2012-11-03T08:57:44Z 2012-11-03T19:19:27Z <p>Can there exist a 'reasonable' extension of the (higher) Chow groups of complex smooth projective algebraic varieties to functors on the category of compact Kähler manifolds? Are there any obstructions for the existence of such a ('nice') extension? In particular, could there exist some 'Chow motives for compact Kähler manifolds'?</p> <p>If one tries to mimick the usual 'algebraic' definitions, then one should define an analogue of algebraic cycles for compact Kähler manifolds. Is it reasonable to consider subsets that are images of compact Kähler manifolds with respect to birational morphisms?</p> <p>Upd. Which (GAGA?) statements could help here? I would be deeply grateful for any references! In particular, does there exist a good exposition of GAGA that includes the following statement: let X and Y be projective complex varieties and let $ϕ: X_h \to Y_h$ be a morphism of analytic spaces, then there is a unique morphism $f : X \to Y$ such that $f_h = ϕ$.</p> http://mathoverflow.net/questions/110273/can-one-ignore-primes-lying-over-l-in-the-fontaine-mazur-conjecture-counterexa Can one ignore primes lying over $l$ in the Fontaine-Mazur conjecture? Counterexamples? Mikhail Bondarko 2012-10-21T22:08:32Z 2012-10-22T01:22:07Z <p>The Fontaine-Mazur conjecture predicts that an $l$-adic Galois representation of a number field is 'geometric' if it is unramified outside a finite set of primes and is De Rham for primes lying over $l$. Now, what happens if one forgets about the latter restriction; are there any counterexamples, and is there any (general?) way to understand that those are not geometric without using the De Rham restriction?</p> http://mathoverflow.net/questions/129377/is-it-possible-to-write-quasiprojective-and-quasifinite-without-a/129378#129378 Comment by Mikhail Bondarko Mikhail Bondarko 2013-05-02T06:05:28Z 2013-05-02T06:05:28Z So, you don't think that &quot;quasi-projective&quot; is a well-established exception? http://mathoverflow.net/questions/128461/repeated-homotopy-category-of-chain-complexes Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-23T17:37:05Z 2013-04-23T17:37:05Z I do not think that considering complexes over $K(C)$ is a good idea; I never met any mention of those. http://mathoverflow.net/questions/75017/decomposition-of-motives-of-cellular-varieties/75020#75020 Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-18T17:10:10Z 2013-04-18T17:10:10Z This argument works in any additive category where we have a similar formula. One can also consider Voevodsky's motives with rational coefficients, or Voevodsky-Suslin finite correspondences with rational coefficients modulo homotopy equivalence here. http://mathoverflow.net/questions/127346/semicontinuity-results-for-weights Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-12T15:57:29Z 2013-04-12T15:57:29Z For weights a-la BBD there is an equality for any proper $f$ (since $f_*=f_!$); see Stabilities 5.1.14. http://mathoverflow.net/questions/127275/what-is-the-purpose-of-section-3-of-bbd/127307#127307 Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-12T07:56:26Z 2013-04-12T07:56:26Z I know that the filtered derived category was used in succeeding papers. Yet I was not able to understand whether it was used in BBD. http://mathoverflow.net/questions/126817/how-would-you-say-that-a-small-category-is-embedded-into-functors-from-a-large-c/126820#126820 Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-08T06:15:52Z 2013-04-08T06:15:52Z Yes, this is quite correct! The category of $R$-modules is quite actual for me; yet I don't want to fix this setting. My problem is: I have a theorem that expresses $C'$ in terms of $C$, and I would also like to express $C$ in terms of functors that its objects induce on $C'$. Yet there are 'too many' functors from $C'$! http://mathoverflow.net/questions/126817/how-would-you-say-that-a-small-category-is-embedded-into-functors-from-a-large-c/126820#126820 Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-08T05:07:01Z 2013-04-08T05:07:01Z Thank you! Sorry; I only just recollected that my $C'$ is isomorphic to the category of all additive functors from $C$ to abelian groups. So, the objects that come from an embedding of $C$ into $C'$ do yield compact generators. Yet are $k$-directed colimits suffice to obtain $C'$ from $C$? This is probably wrong for any $k$. http://mathoverflow.net/questions/126817/how-would-you-say-that-a-small-category-is-embedded-into-functors-from-a-large-c Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-08T04:15:17Z 2013-04-08T04:15:17Z I don't have any reasonable small subcategory inside $C'$. I also recollected that my $C$ does not canonically map into $C'$; I only have a bifunctor $C\times C'\to Ab$. http://mathoverflow.net/questions/126693/singularity-locus-in-terms-of-ideals Comment by Mikhail Bondarko Mikhail Bondarko 2013-04-06T07:23:55Z 2013-04-06T07:23:55Z I would also like to know other possibilities. http://mathoverflow.net/questions/125469/when-r-f-is-regular Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-24T18:27:29Z 2013-03-24T18:27:29Z Yes, but how can I find the regular locus (or some open subscheme in it)? http://mathoverflow.net/questions/125469/when-r-f-is-regular Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-24T17:38:57Z 2013-03-24T17:38:57Z Thank you! Yet I would like to have a finite number of conditions. http://mathoverflow.net/questions/124209/regular-subscheme-of-a-projective-limit-of-schemes Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-16T02:57:01Z 2013-03-16T02:57:01Z Thank you!! It seems that I know how to reduce the general case to the local one. http://mathoverflow.net/questions/124265/non-uniqueness-of-smooth-compactification/124290#124290 Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-12T17:35:14Z 2013-03-12T17:35:14Z So, smooth compactifications are certainly not unique; yet this non-uniqueness can be controlled to a certain extent. http://mathoverflow.net/questions/124265/non-uniqueness-of-smooth-compactification/124290#124290 Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-12T08:59:26Z 2013-03-12T08:59:26Z A remark: though $U$ doesnotdetermine $Y$, for any cohomology theory that factorizes through Voevodsky′s motives the image $H(Y)\to H(U)$ is canonical and functorial (this is the zeroth level of the weight filtration). One can also look at weight complexes. In characteristic $p$ one can prove a somewhat similar result for any cohomology whose target is a $Z[1/p]$-linear category. http://mathoverflow.net/questions/123716/which-schemes-can-be-presented-as-limits-of-smooth-varieties/123719#123719 Comment by Mikhail Bondarko Mikhail Bondarko 2013-03-06T10:07:41Z 2013-03-06T10:07:41Z I'm sorry! I need the connecting morphisms to be dominant, but I forgot to write about this.